Conjugate of an abelian maximal subgroup is maximal

Perhaps this is a trivial question. Let $G$ be a finite group, and $M$ be a maximal (proper) subgroup of $G$. Suppose also that $M$ is abelian. How could I prove that if $x\in G$, then $xMx^{-1}$ is a maximal subgroup of $G$?

Since the conjugation map is bijective, we know that $xMx^{-1}$ has the same size as $M$. But this doesn't necessarily mean $xMx^{-1}$ is a maximal subgroup of $G$, since the inclusion of subgroups is not a total order on the set of all subgroups of $G$. I am not sure what approach one would use to prove $xMx^{-1}$ is not contained in any strictly larger proper subgroup of $G$. Also, is the condition that "$M$ is an abelian" neccesary for the conclusion to hold?

Any help is appreciated! Thanks.

If $H$ is a subgroup of $G$ such that $xMx^{-1} \subsetneq H$, then $M \subsetneq x^{-1}Hx$: what can you deduce about $x^{-1}Hx$? About $H$?
• Oh wow! We would deduce that $x^{-1}Hx$ is the whole group, and so $H$ is the whole group, as desired. Thanks! – Prism Mar 3 '13 at 8:36
• @Prism, so note that you do not need $G$ to be finite, nor $M$ to be abelian for the statement to hold. – Andreas Caranti Mar 4 '13 at 8:46